cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A080510 Triangle read by rows: T(n,k) gives the number of set partitions of {1,...,n} with maximum block length k.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 4, 1, 1, 25, 20, 5, 1, 1, 75, 90, 30, 6, 1, 1, 231, 420, 175, 42, 7, 1, 1, 763, 2016, 1015, 280, 56, 8, 1, 1, 2619, 10024, 6111, 1890, 420, 72, 9, 1, 1, 9495, 51640, 38010, 12978, 3150, 600, 90, 10, 1, 1, 35695, 276980, 244035, 91938, 24024, 4950, 825, 110, 11, 1
Offset: 1

Views

Author

Wouter Meeussen, Mar 22 2003

Keywords

Comments

Row sums are A000110 (Bell numbers). Second column is A001189 (Degree n permutations of order exactly 2).
From Peter Luschny, Mar 09 2009: (Start)
Partition product of Product_{j=0..n-1} ((k + 1)*j - 1) and n! at k = -1, summed over parts with equal biggest part (see the Luschny link).
Underlying partition triangle is A036040.
Same partition product with length statistic is A008277.
Diagonal a(A000217) = A000012.
Row sum is A000110. (End)
From Gary W. Adamson, Feb 24 2011: (Start)
Construct an array in which the n-th row is the partition function G(n,k), where G(n,1),...,G(n,6) = A000012, A000085, A001680, A001681, A110038, A148092, with the first few rows
1, 1, 1, 1, 1, 1, 1, ... = A000012
1, 2, 4, 10, 26, 76, 232, ... = A000085
1, 2, 5, 14, 46, 166, 652, ... = A001680
1, 2, 5, 15, 51, 196, 827, ... = A001681
1, 2 5 15 52 202 869, ... = A110038
1, 2, 5 15 52 203 876, ... = A148092
...
Rows tend to A000110, the Bell numbers. Taking finite differences from the top, then reorienting, we obtain triangle A080510.
The n-th row of the array is the eigensequence of an infinite lower triangular matrix with n diagonals of Pascal's triangle starting from the right and the rest zeros. (End)

Examples

			T(4,3) = 4 since there are 4 set partitions with longest block of length 3: {{1},{2,3,4}}, {{1,3,4},{2}}, {{1,2,3},{4}} and {{1,2,4},{3}}.
Triangle begins:
  1;
  1,    1;
  1,    3,     1;
  1,    9,     4,    1;
  1,   25,    20,    5,    1;
  1,   75,    90,   30,    6,   1;
  1,  231,   420,  175,   42,   7,  1;
  1,  763,  2016, 1015,  280,  56,  8,  1;
  1, 2619, 10024, 6111, 1890, 420, 72,  9,  1;
  ...
		

Crossrefs

Columns k=1..10 give: A000012 (for n>0), A001189, A229245, A229246, A229247, A229248, A229249, A229250, A229251, A229252. - Alois P. Heinz, Sep 17 2013
T(2n,n) gives A276961.
Take differences along rows of A229223. - N. J. A. Sloane, Jan 10 2018

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
           add(b(n-i*j, i-1) *n!/i!^j/(n-i*j)!/j!, j=0..n/i)))
        end:
    T:= (n, k)-> b(n, k) -b(n, k-1):
    seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Apr 20 2012
  • Mathematica
    << DiscreteMath`NewCombinatorica`; Table[Length/@Split[Sort[Max[Length/@# ]&/@SetPartitions[n]]], {n, 12}]
    (* Second program: *)
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, Sum[b[n-i*j, i-1]*n!/i!^j/(n-i*j)!/j!, {j, 0, n/i}]]]; T[n_, k_] := b[n, k]-b[n, k-1]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Feb 25 2014, after Alois P. Heinz *)

Formula

E.g.f. for k-th column: exp(exp(x)*GAMMA(k, x)/(k-1)!-1)*(exp(x^k/k!)-1). - Vladeta Jovovic, Feb 04 2005
From Peter Luschny, Mar 09 2009: (Start)
T(n,0) = [n = 0] (Iverson notation) and for n > 0 and 1 <= m <= n.
T(n,m) = Sum_{a} M(a)|f^a| where a = a_1,...,a_n such that
1*a_1 + 2*a_2 + ... + n*a_n = n and max{a_i} = m, M(a) = n!/(a_1!*...*a_n!),
f^a = (f_1/1!)^a_1*...*(f_n/n!)^a_n and f_n = Product_{j=0..n-1} (-1) = (-1)^n. (End)
From Ludovic Schwob, Jan 15 2022: (Start)
T(2n,n) = C(2n,n)*(A000110(n)-1/2) for n>0.
T(n,m) = C(n,m)*A000110(n-m) for 2m > n > 0. (End)

A364971 Number T(n,k) of partitions of [n] for which the difference between the longest and the shortest block size is k; triangle T(n,k), n>=0, 0<=k<=max(0,n-2), read by rows.

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 6, 4, 2, 35, 10, 5, 27, 60, 95, 15, 6, 2, 371, 315, 161, 21, 7, 142, 938, 2002, 770, 252, 28, 8, 282, 4005, 9744, 5313, 1386, 372, 36, 9, 1073, 16950, 50275, 33705, 11082, 2310, 525, 45, 10, 2, 74657, 283525, 217800, 78078, 20097, 3630, 715, 55, 11
Offset: 0

Views

Author

Alois P. Heinz, Aug 15 2023

Keywords

Comments

T(0,0) = 1 by convention.

Examples

			T(4,0) = 5: 1|2|3|4, 12|34, 13|24, 14|23, 1234.
T(4,1) = 6: 1|2|34, 1|23|4, 1|24|3, 12|3|4, 13|2|4, 14|2|3.
T(4,2) = 4: 1|234, 123|4, 124|3, 134|2.
Triangle T(n,k) begins:
     1;
     1;
     2;
     2,     3;
     5,     6,     4;
     2,    35,    10,     5;
    27,    60,    95,    15,     6;
     2,   371,   315,   161,    21,    7;
   142,   938,  2002,   770,   252,   28,   8;
   282,  4005,  9744,  5313,  1386,  372,  36,  9;
  1073, 16950, 50275, 33705, 11082, 2310, 525, 45, 10;
  ...
		

Crossrefs

Row sums give A000110.
Column k=0 gives A038041 (for n>=1).
T(n,n-2) gives A000027 (for n>=2).

Programs

  • Maple
    b:= proc(n, l, m) option remember; `if`(n=0, x^(m-l), add(
         b(n-j, min(l, j), max(m, j))*binomial(n-1, j-1), j=1..n))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
    seq(T(n), n=0..12);
  • Mathematica
    b[n_, l_, m_] := b[n, l, m] = If[n == 0, x^(m - l), Sum[b[n - j, Min[l, j], Max[m, j]]*Binomial[n - 1, j - 1], {j, 1, n}]];
    T[n_] := CoefficientList[b[n, n, 0], x];
    Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Oct 27 2023, after Alois P. Heinz *)
Showing 1-2 of 2 results.